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- Question
What will be the output of the program?
public class Switch2 { final static short x = 2; public static int y = 0; public static void main(String [] args) { for (int z=0; z < 3; z++) { switch (z) { case x: System.out.print("0 "); case x-1: System.out.print("1 "); case x-2: System.out.print("2 "); } } } }
Options- A. 0 1 2
- B. 0 1 2 1 2 2
- C. 2 1 0 1 0 0
- D. 2 1 2 0 1 2
- Correct Answer
- 2 1 2 0 1 2
ExplanationThe case expressions are all legal because x is marked final, which means the expressions can be evaluated at compile time. In the first iteration of the for loop case x-2 matches, so 2 is printed. In the second iteration, x-1 is matched so 1 and 2 are printed (remember, once a match is found all remaining statements are executed until a break statement is encountered). In the third iteration, x is matched. So 0 1 and 2 are printed.
More questions
- 1. What will be the output of the program?
import java.util.*; public class NewTreeSet2 extends NewTreeSet { public static void main(String [] args) { NewTreeSet2 t = new NewTreeSet2(); t.count(); } } protected class NewTreeSet { void count() { for (int x = 0; x < 7; x++,x++ ) { System.out.print(" " + x); } } }
Options- A. 0 2 4
- B. 0 2 4 6
- C. Compilation fails at line 2
- D. Compilation fails at line 10 Discuss
Correct Answer: Compilation fails at line 10
Explanation:
Nonnested classes cannot be marked protected (or final for that matter), so the compiler will fail at protected class NewTreeSet.- 2. What will be the output of the program?
int x = 3; int y = 1; if (x = y) /* Line 3 */ { System.out.println("x =" + x); }
Options- A. x = 1
- B. x = 3
- C. Compilation fails.
- D. The code runs with no output. Discuss
Correct Answer: Compilation fails.
Explanation:
Line 3 uses an assignment as opposed to comparison. Because of this, the if statement receives an integer value instead of a boolean. And so the compilation fails.- 3. What will be the output of the program?
public static void main(String[] args) { Object obj = new Object() { public int hashCode() { return 42; } }; System.out.println(obj.hashCode()); }
Options- A. 42
- B. Runtime Exception
- C. Compile Error at line 2
- D. Compile Error at line 5 Discuss
Correct Answer: 42
Explanation:
This code is an example of an anonymous inner class. They can be declared to extend another class or implement a single interface. Since they have no name you can not use the "new" keyword on them.In this case the annoynous class is extending the Object class. Within the {} you place the methods you want for that class. After this class has been declared its methods can be used by that object in the usual way e.g. objectname.annoymousClassMethod()
- 4. What will be the output of the program?
public class SyncTest { public static void main (String [] args) { Thread t = new Thread() { Foo f = new Foo(); public void run() { f.increase(20); } }; t.start(); } } class Foo { private int data = 23; public void increase(int amt) { int x = data; data = x + amt; } }and assuming that data must be protected from corruption, what?if anything?can you add to the preceding code to ensure the integrity of data?
Options- A. Synchronize the run method.
- B. Wrap a synchronize(this) around the call to f.increase().
- C. The existing code will cause a runtime exception.
- D. Synchronize the increase() method Discuss
Correct Answer: Synchronize the increase() method
Explanation:
Option D is correct because synchronizing the code that actually does the increase will protect the code from being accessed by more than one thread at a time.Option A is incorrect because synchronizing the run() method would stop other threads from running the run() method (a bad idea) but still would not prevent other threads with other runnables from accessing the increase() method.
Option B is incorrect for virtually the same reason as A?synchronizing the code that calls the increase() method does not prevent other code from calling the increase() method.
- 5. What is the most restrictive access modifier that will allow members of one class to have access to members of another class in the same package?
Options- A. public
- B. abstract
- C. protected
- D. synchronized
- E. default access Discuss
Correct Answer: default access
Explanation:
default access is the "package oriented" access modifier.Option A and C are wrong because public and protected are less restrictive. Option B and D are wrong because abstract and synchronized are not access modifiers.
- 6. Which of the following statements is true?
Options- A. If assertions are compiled into a source file, and if no flags are included at runtime, assertions will execute by default.
- B. As of Java version 1.4, assertion statements are compiled by default.
- C. With the proper use of runtime arguments, it is possible to instruct the VM to disable assertions for a certain class, and to enable assertions for a certain package, at the same time.
- D. When evaluating command-line arguments, the VM gives -ea flags precedence over -da flags. Discuss
Correct Answer: With the proper use of runtime arguments, it is possible to instruct the VM to disable assertions for a certain class, and to enable assertions for a certain package, at the same time.
Explanation:
Option C is true because multiple VM flags can be used on a single invocation of a Java program.Option A is incorrect because at runtime assertions are ignored by default.
Option B is incorrect because as of Java 1.4 you must add the argument -source 1.4 to the command line if you want the compiler to compile assertion statements.
Option D is incorrect because the VM evaluates all assertion flags left to right.
- 7. What will be the output of the program?
public class Switch2 { final static short x = 2; public static int y = 0; public static void main(String [] args) { for (int z=0; z < 3; z++) { switch (z) { case x: System.out.print("0 "); case x-1: System.out.print("1 "); case x-2: System.out.print("2 "); } } } }
Options- A. 0 1 2
- B. 0 1 2 1 2 2
- C. 2 1 0 1 0 0
- D. 2 1 2 0 1 2 Discuss
Correct Answer: 2 1 2 0 1 2
Explanation:
The case expressions are all legal because x is marked final, which means the expressions can be evaluated at compile time. In the first iteration of the for loop case x-2 matches, so 2 is printed. In the second iteration, x-1 is matched so 1 and 2 are printed (remember, once a match is found all remaining statements are executed until a break statement is encountered). In the third iteration, x is matched. So 0 1 and 2 are printed.- 8. Which is the valid declarations within an interface definition?
Options- A. public double methoda();
- B. public final double methoda();
- C. static void methoda(double d1);
- D. protected void methoda(double d1); Discuss
Correct Answer: public double methoda();
Explanation:
Option A is correct. A public access modifier is acceptable. The method prototypes in an interface are all abstract by virtue of their declaration, and should not be declared abstract.Option B is wrong. The final modifier means that this method cannot be constructed in a subclass. A final method cannot be abstract.
Option C is wrong. static is concerned with the class and not an instance.
Option D is wrong. protected is not permitted when declaring a method of an interface. See information below.
Member declarations in an interface disallow the use of some declaration modifiers; you cannot use transient, volatile, or synchronized in a member declaration in an interface. Also, you may not use the private and protected specifiers when declaring members of an interface.
- 9. Suppose that you would like to create an instance of a new Map that has an iteration order that is the same as the iteration order of an existing instance of a Map. Which concrete implementation of the Map interface should be used for the new instance?
Options- A. TreeMap
- B. HashMap
- C. LinkedHashMap
- D. The answer depends on the implementation of the existing instance. Discuss
Correct Answer: LinkedHashMap
Explanation:
The iteration order of a Collection is the order in which an iterator moves through the elements of the Collection. The iteration order of a LinkedHashMap is determined by the order in which elements are inserted.When a new LinkedHashMap is created by passing a reference to an existing Collection to the constructor of a LinkedHashMap the Collection.addAll method will ultimately be invoked.
The addAll method uses an iterator to the existing Collection to iterate through the elements of the existing Collection and add each to the instance of the new LinkedHashMap.
Since the iteration order of the LinkedHashMap is determined by the order of insertion, the iteration order of the new LinkedHashMap must be the same as the interation order of the old Collection.
- 10. Which is a valid declarations of a String?
Options- A. String s1 = null;
- B. String s2 = 'null';
- C. String s3 = (String) 'abc';
- D. String s4 = (String) '\ufeed'; Discuss
Correct Answer: String s1 = null;
Explanation:
Option A sets the String reference to null.Option B is wrong because null cannot be in single quotes.
Option C is wrong because there are multiple characters between the single quotes ('abc').
Option D is wrong because you can't cast a char (primitive) to a String (object).
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- 1. What will be the output of the program?